Obtaining the fitted values of main effects and interactions in PROC GLM or MIXED

[JamesLin]

In short, I need to decompose the fitted values obtained from ANOVA into components corresponding to each term in ANOVA statement. This problem appears trivial, but it becomes tricky when GLM coding is used for the design matrix. Let's take a balanced 2*2 ANOVA with 2 observations per cell.

Y = A + B + A*B

GLM coding produces 8*9 design matrix:

Intercept – column 0

Main effect of A – columns 1-2

Main effect of B – columns 3-4

Interaction – columns 5-8

The estimated coefficients, b, look like (4 estimable parameters, as expected):

b0 b1 b2 b3 b4 b5 b6 b7 b8

5.5 -0.915 0 1.15 0 -0.36 0 0 0

Let's take the covariate pattern:

1 1 0 1 0 1 0 0 0

for which the fitted value is 5.375. One could say that it's decomposed as

5.375 = 5.5 (Int) - 0.915 (A main effect) + 1.15 (B main effect) - 0.36 (A*B interaction)

but that's not the case. The linear restrictions for Type III hypotheses to test the significance of the two main effects and interactions are created as specified here.  In particular, to test

H0: Main effect of A = 0,

we test whether k′b where k is (defined up to a multiple):

0 0.894427 -0.894427 0 0 0.447214 0.447214 -0.447214 -0.447214

That shows that, for the given covariate pattern, the point estimate of A main effect depends on both b1 and b5, not just b1.

To test H0: Main effect of B = 0, we use the restriction:

0 0 0 0.894427 -0.894427 0.447214 -0.447214 0.447214 -0.447214

and for H0: A*B = 0 it is:

0 0 0 0 0 1 -1 -1 1

These linear restrictions are called "Type III estimable functions" in PROC GLM and "Type III coefficients" in PROC MIXED. The problem is I don't know how to use all that information in order to obtain the point estimates of main effects and interactions for a given covariate pattern.

[PaigeMiller]

The estimated coefficients, b, look like (4 estimable parameters, as expected):

b0 b1 b2 b3 b4 b5 b6 b7 b8

5.5 -0.915 0 1.15 0 -0.36 0 0 0

Let's take the covariate pattern:

1 1 0 1 0 1 0 0 0

for which the fitted value is 5.375. One could say that it's decomposed as

5.375 = 5.5 (Int) - 0.915 (A main effect) + 1.15 (B main effect) - 0.36 (A*B interaction)

but that's not the case.

I'm afraid that is the case. If you want a predicted value of Y for this X condition, you have done the right thing.

The linear restrictions for Type III hypotheses to test the significance of the two main effects and interactions are created as specified here.  In particular, to test

H0: Main effect of A = 0,

we test whether k′b where k is (defined up to a multiple):

0 0.894427 -0.894427 0 0 0.447214 0.447214 -0.447214 -0.447214

The linear restrictions to test the main effect of A have nothing to do with prediction of a value at a given X condition, as you discussed above. They are two different things. You are mixing apples and gorillas.

[JamesLin]

Thanks for replying. To put it in other terms, I want to decompose the fitted value into two main effects and interaction term to be able to adjust the observed response. E.g. if I want to adjust it for B and A*B, 

Y_adj = Y - (Main effect of B) - (AB interaction)

so that when I regress Y_adj on A, B, A*B again, the Type III p-value for A should be exactly the same, and the p-values for B and A*B should be equal to one. I tried it the naive way (e.g. for that observation add (1.15 - 0.36)  to adjust it), but the p-value for A is not preserved.

[PaigeMiller]

---

@JamesLin wrote:

E.g. if I want to adjust it for B and A*B, 

Y_adj = Y - (Main effect of B) - (AB interaction)

so that when I regress Y_adj on A, B, A*B again, the Type III p-value for A should be exactly the same

I doubt this is possible

and the p-values for B and A*B should be equal to one

In combination, these two conditions seems like a very unusual way of doing things, and I doubt it is possible.

[PaigeMiller]

Adding: the p-value depends on the number of degrees of freedom of the error term, and the estimate of the root mean square error of the model. As soon as you try to fix one portion of the model, and let other parts vary, you have changed the estimate of the root mean square error of the model, and so you cannot get the same p-values.

[sld]

The estimated coefficients look like REF coding, not GLM coding. Some PROCs (e.g., LOGISTIC, GLMSELECT) have options for other CLASS parameterizations, but I think that the GLM and MIXED procedures use only REF by default. If so, your confusion may be due to thinking you have GLM parameterization but you actually have REF.

Posting your code and an example dataset could be quite helpful.

Also, I might think that "the point estimates of main effects and interactions for a given covariate pattern" are just the interaction LSMEANS. It's not clear to me what you are trying to obtain.

[JamesLin]

No, it's not the reference coding. One can call it "Intercept + GLM"; e.g. for Y = A, the design has a column of 1's plus k columns where k is is the number of levels of factor A. If it were reference or effect coding the design would have fewer than 9 columns. I attached the data and code. Could you elaborate what you said about lsmeans?

FactorA FactorB Response
A1 B1 4.48
A1 B2 5.53
A1 B1 4.69
A1 B2 5.22
A2 B1 5.54
A2 B2 6.4
A2 B1 5.46
A2 B2 6.9

proc glm (or mixed) data = input.data;
class FactorA FactorB;
model Response = FactorA FactorB FactorA * FactorB / e e1 e2 e3 e4;
run;

[sld]

Oh, yes, you're right, it is GLM parameterization.

These ESTIMATE statements duplicate the hypothesis tests reported in the Type III ANOVA table. The coefficients are extracted from the Type III Estimable Functions table.

data test;
input FactorA $ FactorB $ Response;
datalines;
A1 B1 4.48
A1 B2 5.53
A1 B1 4.69
A1 B2 5.22
A2 B1 5.54
A2 B2 6.4
A2 B1 5.46
A2 B2 6.9
;
run;
 
proc glm data =test;
class FactorA FactorB;
model Response = FactorA FactorB FactorA * FactorB / e3 solution;
lsmeans FactorA*FactorB;
estimate "Main effect FactorA" intercept 0 FactorA 1 -1 FactorB 0 0 FactorA*FactorB 0.5 0.5 -0.5 -0.5;
estimate "Main effect FactorB" intercept 0 FactorA 0 0 FactorB 1 -1 FactorA*FactorB 0.5 -0.5 0.5 -0.5;
estimate "Interaction" intercept 0 FactorA 0 0 FactorB 0 0 FactorA*FactorB 1 -1 -1 1;
run;

The interaction lsmeans depict the predicted value for each of the 2 x 2 = 4 covariate patterns (i.e., A1B1, A1B2, A2B1, A2B2).

[sld · Accepted solution]

These ESTIMATE statements are linear contrasts for the means for each covariate pattern, and match the LSMEAN output for FactorA*FactorB:

estimate "A1B1" intercept 1 FactorA 1 0 FactorB 1 0 FactorA*FactorB 1 0 0 0;
estimate "A1B2" intercept 1 FactorA 1 0 FactorB 0 1 FactorA*FactorB 0 1 0 0;
estimate "A2B1" intercept 1 FactorA 0 1 FactorB 1 0 FactorA*FactorB 0 0 1 0;
estimate "A2B2" intercept 1 FactorA 0 1 FactorB 0 1 FactorA*FactorB 0 0 0 1;

[JamesLin]

Thanks for clarifying things for me.